IMSA Probability and Combinatorics 2 — Questions and Answers
Question 1: How many ways can you choose 3 students from a class of 10 (order doesn't matter)?
- 120 (Correct answer)
- 720
- 30
- 360
Correct answer: 120
C(10,3) = 10!/(3! × 7!) = (10 × 9 × 8)/(3 × 2 × 1) = 120.
Question 2: Two dice are rolled. What is the probability that both show a 6?
- 1/36 (Correct answer)
- 1/6
- 2/36
- 1/12
Correct answer: 1/36
P(6) × P(6) = (1/6) × (1/6) = 1/36.
Question 3: A spinner has 8 equal sections numbered 1−8. What is the probability of landing on an odd number?
- 1/2 (Correct answer)
- 3/8
- 5/8
- 1/4
Correct answer: 1/2
The odd numbers are 1, 3, 5, 7 (four values), so P = 4/8 = 1/2.
Question 4: How many different 3-letter arrangements can be made from the letters A, B, C, D (no repetition)?
- 24 (Correct answer)
- 12
- 64
- 6
Correct answer: 24
This is a permutation: P(4,3) = 4 × 3 × 2 = 24.
Question 5: Events A and B are mutually exclusive. P(A) = 0.3 and P(B) = 0.4. What is P(A or B)?
- 0.7 (Correct answer)
- 0.12
- 0.1
- 1.0
Correct answer: 0.7
For mutually exclusive events, P(A or B) = P(A) + P(B) = 0.3 + 0.4 = 0.7.
Question 6: A card is drawn from a standard deck. What is the probability it is a heart or a king?
- 4/13 (Correct answer)
- 17/52
- 16/52
- 5/13
Correct answer: 4/13
P(heart) = 13/52, P(king) = 4/52, P(king of hearts) = 1/52; P(heart or king) = 16/52 = 4/13.
IMSA Probability and Combinatorics 2
How many ways can you choose 3 students from a class of 10 (order doesn't matter)?